Finding character at particular location of String

Question

As a part of a Binomial expansion program, I created a Scanner class (sc) on an entered number + variable, and another (scan)on the first term (a in (a+b)ⁿ)

Here's my program till now :

import java.util.* ;
public class binomial_expansion
{
    public static long factorial(long x)
    {
        long xf = x ;
        for(long i = 1; i<=x ; i++)
        {
            xf*=i ;
        }
        return xf ;
    }
    
    public static double C(long n, long r)
    {
        double c = factorial(n)/(factorial(n-r)*factorial(r)) ;
        return c ;
    }
    
    public static void main()
    {
        Scanner sc = new Scanner(System.in) ;
        boolean error = false ;
        String a = "" ;
        do {
            if(error)
            {
                System.out.println("Error; Please try again") ;
                error = false ;
            }
            System.out.println("Enter a (separate single variable with space) : ") ;
            a = sc.nextLine() ;
            if(a=="") error = true ;
        } while(error) ;
        
        sc.close() ;
        Scanner scan = new Scanner(a) ;
        long x = scan.nextLong() ;
        long x1 = scan.next().charAt(0) ;
    }
}

The problem with creating a character variable using `.next().charAt(index)` is that the index number of the character (the single variable) is not known in the string (the length of the integer is not known), and also that the blank space has to be omitted which was entered by the user to separate the integer literal and the character literal. How can the character variable containing the single-digit variable be made ?

I don't understand the purpose of your `error` variable which is only ever set to false. — khelwood, May 17 '23 at 18:11
Save the input to a string, use a regular expression to parse it and capture the individual elements into groups. What exact input do you expect? — f1sh, May 17 '23 at 18:17
To clarify your question, you're requesting from the user to enter just 1 character and 1 number? — Reilas, May 17 '23 at 18:17
First problem, you're using the parameter `x` as a limit in `for(long i = 1; i<=x ; i++)`, but within the loop you're changing `x*=i`. `x` will increase faster than `i`, and the loop won't work. Use another variable (e.g. `var xf = x`) for the return value, and declare your parameter as `final long x` to ensure it's not changed. — John Bayko, May 17 '23 at 18:19
@Reilas for now, only one number, space, and a single digit veriable — Super Coder, May 17 '23 at 18:20
`error` isn't being used correctly. Get the base case working first, then add error handling. — John Bayko, May 17 '23 at 18:22
You should initialize xf to be $1$ in your factorial program. It looks like this will return $(x!)x$ — John Douma, Jun 07 '23 at 20:39

JustaNobody · Answer 1 · 2023-05-17T18:31:48.583

1

You can read a string and then use stringtokenizer to seperate with black spaces. Then use nextToken() to get the integer and variable seperately.

Long.parseLong() to convert string to long

import java.util.*;
class test{
    public static void main(String args[]){
        String str = "12 w";
        StringTokenizer st = new StringTokenizer(str," ");
        int x = Integer.parseInt(st.nextToken());
        str=st.nextToken();
        System.out.println(str+"-"+x);
    }
}

Here I used int, you can use long aswell

edited May 17 '23 at 18:31

answered May 17 '23 at 18:23

JustaNobody

150
8

1

yes StringTokeniser class can be used here as well as sc.substring – Super Coder May 17 '23 at 18:34
2

I would *prefer* a `Scanner` instead of `StringTokenizer` - it separates by whitespaces and also can parse integer (and is a bit newer, or less older) – user16320675 May 18 '23 at 05:29
@user16320675 Yeah using delimiter would be good for replacing stringtokenizer – JustaNobody May 18 '23 at 17:17

Reilas · Accepted Answer · 2023-05-17T22:36:21.637

1

You'll only need one Scanner to get the input, which you do with a = sc.nextLine().

From there, find the index of the space, and use the String.substring method to derive each value.

Scanner sc = new Scanner(System.in) ;
boolean error = false ;
String a = "" ;
do {
    if(error)
    {
        System.out.println("Error; Please try again") ;
        error = false ;
    }
    System.out.println("Enter a (separate single variable with space) : ") ;
    a = sc.nextLine() ;
} while(error) ;

sc.close() ;

int indexOf = a.indexOf(' ');
long number = Long.parseLong(a.substring(0, indexOf));
char variable = a.charAt(indexOf + 1);

edited May 17 '23 at 22:36

answered May 17 '23 at 18:26

Reilas

3,297
2
4
17

1

`a.substring(indexOf + 1).charAt(0)` would be the same as `a.charAt(indexOf + 1)` - just more method call and intermediate `String` created || In *production* code you would also check if `indexOf` is not `-1`... – user16320675 May 17 '23 at 20:53
@user16320675, you're right, I overlooked that. I made the change to my answer. – Reilas May 17 '23 at 22:37

Finding character at particular location of String

2 Answers2